Solve functional equation $(x+y)(f(x)-f(y))=f(x^2)-f(y^2)$ [duplicate]

Find all real functions $f\colon \mathbb{R}\rightarrow \mathbb{R}$ so that $(x+y)(f(x)-f(y))=f(x^2)-f(y^2)$.

Can someone at least find the value of $f(1)$ if it is possible, it would help me.

Solution 1:

Without loss of generality, lets look for solutions to the equation satisfying $f(0)=0$. The equation then becomes $$xf(x)=f(x^2),\ x\in \mathbb R. $$ Therefore $$(x+y)(f(x)-f(y))=f(x^2)-f(y^2)=xf(x)-yf(y),$$ from which we get (by expanding the LHS) $$yf(x)=xf(y) $$ for all $x,y\in \mathbb R$. Setting $y=1$ we see that all solutions are of the form $$f(x)=kx.$$

Solution 2:

I made a mistake last time. Now everything should be fine.

Putting $y=0$ we get $f(x^2)=x(f(x)-f(0))+f(0)$. Putting that in the first equation leaves us with $(x+y)(f(x)-f(y))=x(f(x)-f(0))+f(0)-y(f(y)-f(0))-f(0)$

$y f(x)-x f(y)=-f(0)(x-y)$

With $y=1$ we get: $f(x)-x f(1)=-f(0)(x-1) \Rightarrow f(x)=x(f(1)-f(0))+f(0)$ If we put $f(1)-f(0)=a$ and $f(0)=b$ and then we can write: $f(x)=ax+b$.